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(D) For compound $P$ ($4$-hydroxybenzoic acid): The $-OH$ group is strongly activating and ortho/para-directing,while the $-COOH$ group is deactivatin

Vedclass · Accepted Answer

For $P$: $3$-nitro$-4-$hydroxybenzoic acid; For $Q$: $2$-nitro$-1-$methoxy$-4-$methylbenzene; For $S$: $3$-nitrophenyl benzoate

Vedclass · Answer

(D) For compound $P$ ($4$-hydroxybenzoic acid): The $-OH$ group is strongly activating and ortho/para-directing,while the $-COOH$ group is deactivating and meta-directing. The $-OH$ group dominates,directing the nitro group to the ortho position relative to itself (position $3$ relative to $-COOH$).For compound $Q$ ($1$-methoxy$-4-$methylbenzene): The $-OCH_3$ group is strongly activating and ortho/para-directing,while the $-CH_3$ group is weakly activating and ortho/para-directing. The $-OCH_3$ group dominates,directing the nitro group to its ortho position (position $2$ relative to $-OCH_3$).For compound $S$ (phenyl benzoate): The benzoate group $(-O-CO-C_6H_5)$ is deactivating and meta-directing. However,the ring attached to the oxygen atom is activated by the oxygen lone pair (ortho/para-directing). Nitration occurs on the ring attached to the oxygen,specifically at the meta position relative to the ester linkage due to the electronic influence of the ester group on the phenoxy ring. The correct product is $3$-nitrophenyl benzoate.

What are the major products obtained in each case when compounds $P, Q,$ and $S$ are nitrated using a mixture of $HNO_3/H_2SO_4$?

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